I'm reading some Galois theory in Lang's Algebra, and he often refers to maps acting on elements of a field extension as embeddings in the algebraic closure of the base field (if I'm not mistaken). However, it seems that the things he refers to as embeddings are replaced by elements of the Galois group in many other texts, for instance in the definition of norm and trace. Can someone tell me how these two things are related? I guess I can see it intuitively, but why when the extension is Galois do embeddings into the algebraic closure become automorphisms? Any help would be much appreciated.



Any embedding of the extension into the algebraic closure must take each element of the extension to one of its conjugates over the base field. Part of the definition of Galois extension is that it contains all the conjugates of each of its elements. So for a Galois extension, an embedding is an automorphism.

  • $\begingroup$ Why must an embedding take an element to a conjugate? $\endgroup$ – Jonathan Beardsley Apr 20 '11 at 3:12
  • 1
    $\begingroup$ Let $q$ be in the extension. Let $f$ be the monic irreducible polynomial it satisfies over the base field. Let $T$ be the embedding. $T$ is a field homomorphism, and is the identity on the base field, so $0=T(0)=T(f(q))=f(T(q))$. Thus, $T(q)$ is a conjugate of $q$. $\endgroup$ – Gerry Myerson Apr 20 '11 at 3:31
  • $\begingroup$ Oh that's simple. Thanks so much. $\endgroup$ – Jonathan Beardsley Apr 20 '11 at 3:41
  • 6
    $\begingroup$ In the interest of precision, I would prefer that you say "for a Galois extension, any two embeddings are related by an automorphism (of the image of the embedding)." $\endgroup$ – Qiaochu Yuan Apr 20 '11 at 4:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.